Answer:
The answer is 465.6 mg of MgI₂ to be added.
Explanation:
We find the mole of ion I⁻ in the final solution
C = n/V -> n = C x V = 0.2577 (L) x 0.1 (mol/L) = 0.02577 mol
But in the initial solution, there was 0.087 M KI, which can be converted into mole same as above calculation, equal to 0.02242 mol.
So we need to add an addition amount of 0.02577 - 0.02242 = 0.00335 mol of I⁻. But each molecule of MgI₂ yields two ions of I⁻, so we need to divide 0.00335 by 2 to find the mole of MgI₂, which then is 0.001675 mol.
Hence, the weight of MgI₂ must be added is
Weight of MgI₂ = 0.001675 mol x 278 g/mol = 0.4656 g = 465.6 mg
Answer:
it's food engineering obviously
The density of CO2 getting from experiment is 0.1/0.056 = 1.79 g/L. The percent error of this is (1.96 -1.79)/1.96*100%=8.67%. So the approximate percent error is 8.67%.
Answer:
Alkanes
Explanation:
Alkanes are hydrocarbons containing only single bonds between the parent chain carbons.
Alkenes are hydrocarbons containing at least one double bond between the parent chain carbons.
Alkynes are hydrocarbons containing at least one triple bond between the parent chain carbons.
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